hwsol4 - Homework 5 Solutions Course Reader Exercises 15.3...

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Unformatted text preview: Homework 5 Solutions Course Reader Exercises 15.3 The matrix for the rotation through / 4 radians is A = " 2 2- 2 2 2 2 2 2 # Since the reflection sends e 1 to- e 2 and e 2 to- e 1 , its matrix is B =- 1- 1 Since T is the composition of these transformations, its matrix is BA = "- 2 2- 2 2- 2 2 2 2 # 15.4 First we need to consider the effect T has on the standard basis vectors in R 3 . Since the x-axis is fixed by T , T ( e 1 ) = e 1 . The figure below illustrates the effect T has on the y- z plane. y z e e e e T( ) T( ) 2 2 3 3 The x components of both T ( e 2 ) and T ( e 3 ) are zero. Using a little trigonometry, the y and z components of T ( e 2 ) are seen to be 3 / 2 and 1 / 2, respectively. Likewise the y and z components of T ( e 3 ) are- 1 / 2 and 3 / 2, respectively. Thus T ( e 1 ) = 1 T ( e 2 ) = 3 / 2 1 / 2 T ( e 3 ) = - 1 / 2 3 / 2 so the matrix for T is A = 1 3 / 2- 1 / 2 1 / 2 3 / 2 1 Since S leaves the z-axis fixed, S ( e 3 ) = e 3 . The effect of S on the x- y plane is illustrated below. x y e e e e T( ) T( ) 1 1 2 2 The z components of both S ( e 1 ) and S ( e 2 ) are zero. The x and y components of S ( e 1 ) are both 2 / 2 and the x and y components of S ( e 2 ) are- 2 / 2 and 2 / 2, respectively. Thus S ( e 1 ) = 2 / 2 2 / 2 S ( e 2 ) = - 2 / 2 2 / 2 S ( e 3 ) = 1 so the matrix for S is B = 2 / 2- 2 / 2 0 2 / 2 2 / 2 1 The matrix for T S is AB = 2 / 2- 2 / 2 6 / 4 6 / 4- 1 / 2 2 / 4 2 / 4 3 / 2 The matrix for S T is BA = 2 / 2- 6 / 4 2 / 4 2 / 2 6 / 4- 2 / 4 1 / 2 3 / 2 16.11 Remember that I n A = AI n = A . Now expand ( I n + A )( I n- A ) = I n I n- I n A + AI n- AA = I n- AA = I n- A 2 16.12 Be careful. AB 6 = BA ! So we expand ( B + A )( B- A ) = BB- BA + AB- AA 2 16.13 Saying that A = A- 1 is the same as saying that AA = I : a b b- a a b b- a = 1 0 0 1 a 2 + b 2 a 2 + b 2 = 1 0 0 1 So we deduce that a 2 + b 2 = 1. 16.14 For the matrix to have an inverse, its reduced row echelon from must be the identity there should be no free variables. So lets try and row reduce as best as we can. R 2- R 1 1 1 1 0 1 k- 1 1 4 k 2 R 3- R 1 1 1 1 0 1 k- 1 0 3 k 2 R 1- R 2 1 0 2- k 0 1 k- 1 0 3 k 2 R 3- 3 R 2 1 0 2- k 0 1 k- 1 0 0 k 2- 3 k + 2 Now we can ensure that the first two columns are pivot columns. The third column will be a pivot column provided that k 2- 3 k + 2 6 = 0 in which case we can use it to eliminate the other entries in the third column. More geometrically, provided that k 2- 3 k + 2 6 = 0, the vectors corresponding to our three columns will be linearly independent....
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This note was uploaded on 03/01/2008 for the course MATH 51 taught by Professor Staff during the Spring '07 term at Stanford.

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hwsol4 - Homework 5 Solutions Course Reader Exercises 15.3...

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